UVA-100
2017-08-23 15:34
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Problems in Computer Science are often classified as belonging to a certain class of problems (e.g.,
NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose
classification is not known for all possible inputs.
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then n −3n + 1
5. else n −n/2
6. GOTO 2
Given the input 22, the following sequence of numbers will be printed
22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input
value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has
been verified, however, for all integers n such that 0 < n < 1; 000; 000 (and, in fact, for many more
numbers than this.)
Given an input n, it is possible to determine the number of numbers printed before and including
the 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cycle
length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers
between and including both i and j .
Input
The input will consist of a series of pairs of integers i and j , one pair of integers per line. All integers
will be less than 10,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over
all integers between and including i and j .
You can assume that no operation overflows a 32-bit integer.
Output
For each pair of input integers i and j you should output i, j , and the maximum cycle length for
integers between and including i and j . These three numbers should be separated by at least one space
with all three numbers on one line and with one line of output for each line of input. The integers i
and j must appear in the output in the same order in which they appeared in the input and should be
followed by the maximum cycle length (on the same line).
Sample Input
1 10
100 200
201 210
900 1000
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174
题意:给出区间数, i 和 j, 找出其中的最大循环节长度(给定一个数,如果他是奇数就乘3加1, 如果他是偶数就除以2,一直到该数成为1为止)。
题解:该题的数据有点水, 暴力可以过,但是要注意给定的i 和 j , i 和 j 的大小不确定, 最后输出的时候也要注意按照输入的顺序输出。
但是暴力最极端的例子, 是从1到最大, 这时候暴力实际上是超时的, 看了别人的思路, 自己原来不知道怎么打表, 就像备忘录(记忆化搜索)一样。打出表。
代码:
#include <iostream>
#include<algorithm>
#include<stdlib.h>
#include<cstdio>
#include<string.h>
using namespace std;
int number[1000000];
int caculator(long long t) {
if(t == 1)
return 1;
if(t & 1) {
t += (t << 1) + 1;
}
else
t >>= 1;
if(t < 1000000) {
if(!number[t])
number[t] = caculator(t);
return 1 + number[t];
}
return 1 + caculator(t);
}
int main() {
int a, b, t1, t2, t3, Max;
memset(number, 0, sizeof(number));
for(int i = 1; i < 1000000; i++) {
number[i] = caculator(i);
}
while(scanf("%d%d", &a, &b) != EOF) {
Max = 0;
t1 = a + b;
t2 = a; t3 = b;
a = min(a, b);
b = t1 - a;
for(int i = a; i <= b; i++)
if(number[i] > Max)
Max = number[i];
printf("%d %d %d\n", t2, t3, Max);
}
}
NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose
classification is not known for all possible inputs.
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then n −3n + 1
5. else n −n/2
6. GOTO 2
Given the input 22, the following sequence of numbers will be printed
22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input
value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has
been verified, however, for all integers n such that 0 < n < 1; 000; 000 (and, in fact, for many more
numbers than this.)
Given an input n, it is possible to determine the number of numbers printed before and including
the 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cycle
length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers
between and including both i and j .
Input
The input will consist of a series of pairs of integers i and j , one pair of integers per line. All integers
will be less than 10,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over
all integers between and including i and j .
You can assume that no operation overflows a 32-bit integer.
Output
For each pair of input integers i and j you should output i, j , and the maximum cycle length for
integers between and including i and j . These three numbers should be separated by at least one space
with all three numbers on one line and with one line of output for each line of input. The integers i
and j must appear in the output in the same order in which they appeared in the input and should be
followed by the maximum cycle length (on the same line).
Sample Input
1 10
100 200
201 210
900 1000
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174
题意:给出区间数, i 和 j, 找出其中的最大循环节长度(给定一个数,如果他是奇数就乘3加1, 如果他是偶数就除以2,一直到该数成为1为止)。
题解:该题的数据有点水, 暴力可以过,但是要注意给定的i 和 j , i 和 j 的大小不确定, 最后输出的时候也要注意按照输入的顺序输出。
但是暴力最极端的例子, 是从1到最大, 这时候暴力实际上是超时的, 看了别人的思路, 自己原来不知道怎么打表, 就像备忘录(记忆化搜索)一样。打出表。
代码:
#include <iostream>
#include<algorithm>
#include<stdlib.h>
#include<cstdio>
#include<string.h>
using namespace std;
int number[1000000];
int caculator(long long t) {
if(t == 1)
return 1;
if(t & 1) {
t += (t << 1) + 1;
}
else
t >>= 1;
if(t < 1000000) {
if(!number[t])
number[t] = caculator(t);
return 1 + number[t];
}
return 1 + caculator(t);
}
int main() {
int a, b, t1, t2, t3, Max;
memset(number, 0, sizeof(number));
for(int i = 1; i < 1000000; i++) {
number[i] = caculator(i);
}
while(scanf("%d%d", &a, &b) != EOF) {
Max = 0;
t1 = a + b;
t2 = a; t3 = b;
a = min(a, b);
b = t1 - a;
for(int i = a; i <= b; i++)
if(number[i] > Max)
Max = number[i];
printf("%d %d %d\n", t2, t3, Max);
}
}
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